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For a while I've been reading J.E.Humphreys's book "Representations of semisimple Lie algebras in the BGG category $\mathcal O$" under the impression that any module in $\mathcal O$ has a finite generating set composed of highest weight vectors. Now I've realised that I'm lacking a proof. So what would serve as a counterexample? Or has my assumption been correct for some reason?

I apologize if the answer can be found further on in the book itself!

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2 Answers 2

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This is false. Consider the contragradient dual on a Verma module $V_\lambda$. This is a module given by linear functions on $V_\lambda$ which kill all but finitely many weight spaces. This module has no highest weight vectors of weight other than $\lambda$; any non-zero vector $\xi$ of weight $< \lambda$ is non-zero on $F_i v$ for some $v$ (since $V_\lambda$ is generated in degree $\lambda$), and $E_i\xi\neq 0$ since it has non-zero value on $v$.

Thus, this module is generated by weight vectors if and only if it is generated by vectors of weight $\lambda$, but this is only possible if the Verma module is irreducible (otherwise, the dual of the simple quotient of the Verma is the proper submodule generated by the vectors of degree $\lambda$). Thus, this is a counter-example.

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I'm somewhat confused. Isn't the set of weights $\Pi(V_\lambda^*)$ equal to $-\Pi(V_\lambda)$? This would stop $V_\lambda^*$ from lying in $\mathcal O$ since its set of weights would have no maximal element. –  Igor Makhlin Aug 28 '12 at 23:18
@Igor: Ben isn't talking about the usual dual space, but rather the BGG dual module. See my clarification and example. –  Jim Humphreys Aug 28 '12 at 23:28
Yes, just before you posted I'd understood it was this modified concept of duality I wasn't grasping. Will have to read up on that! –  Igor Makhlin Aug 28 '12 at 23:32

Ben's answer is essentially correct, modulo his invented word "contragradient" (which like "indeterminant" I've been trying to stamp out, without success). Of course, the definition shows that all modules in the category are finitely generated, so the problem involves generation by highest weight vectors.

Maybe I should comment further on the simplest counterexample, which occurs already in rank 1. It takes a while to build up complicated examples in category $\mathcal{O}$, since I only introduce the BGG duality in Chapter 3. While the Lie algebra acts naturally on the usual vector space dual of a module, this dual is usually too big to lie in $\mathcal{O}$, so as Ben suggests there is a more restrictive notion of duality; this stays in the caegory and preserves the formal characters.

For the rank 1 simple Lie algebra, integral weights may be identified with ordinary integers. In particular, you get a Verma module $M(0)$ with only two composition factors: $L(0)$ is the one-dimensional module at the top, while $L(-2) = M(-2)$ is the unique maximal submodule. Of course, any Verma module is generated by a highest weight vector. But the BGG dual $M(0)^\vee$ has the same two composition factors in reverse order and in particular can't be generated by highest weight vectors since $-2$ isn't a highest weight vector here (see the discussion of Hom in my section 3.3).

In higher ranks you start to run into more sophisticated examples where the maximal submodule of a (non-dominant) Verma module is itself not generated by its highest weight vectors. Indeed, a Verma module might have infinitely many distinct submodules. This was first appreciated by BGG and Conze-Duflo, but is best understood via the later Kazhdan-Lusztig theory.

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I'll note, there's a difference between inventing words and misspelling them. –  Ben Webster Aug 29 '12 at 3:26
@Ben: True. My real point is that "contragredient" is usually just a synonym for "dual" in the context of dual spaces of vector spaces. I guess some people might want to apply the term "contravariant" (derived from "contravariant form") to the BGG dual. Still, "contragradient" or "contragradiant" is one of those odd inventions which doesn't seem to work. –  Jim Humphreys Aug 29 '12 at 12:02

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